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2 years ago

1. The half-life of Co-60 is 5.3 years. If the initial mass of the sample is 1000g, how much will remain after 15.9 years.

Chemistry

1 answer:

mixer [17]2 years ago

8 0

Answer:

125 g

Explanation:

The following data were obtained from the question:

Half-life (t½) = 5.3 years.

Original amount (N₀) = 1000 g

Time (t) = 15.9 years.

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 5.3 years.

Time (t) = 15.9 years.

Number of half-lives (n) =?

n = t / t½

n = 15.9 / 5.3

n = 3

Thus, 3 half-lives has elapsed.

Finally, we shall determine the amount of Co–60 that is remaining. This can be obtained as follow:

Original amount (N₀) = 1000 g

Number of half-lives (n) = 3

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2³ × 1000

N = 1/8 × 1000

N = 0.125 × 1000

N = 125 g

Therefore, the amount of Co–60 that is remaining is 125 g

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Arbon dioxide is dissolved in blood (ph 7.5) to form a mixture of carbonic acid and bicarbonate. Part a neglecting free co2, wha

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Answer : The fraction of carbonic acid present in the blood is 5.95%

Explanation :

The mixture consists of carbonic acid ( H₂CO₃) and bicarbonate ion ( HCO₃⁻). This represents a mixture of weak acid and its conjugate which is a buffer.

The pH of a buffer is calculated using Henderson equation which is given below.

$pH = pKa + log \frac{[Base]}{[Acid]}$

We have been given,

pH = 7.5

pKa of carbonic acid = 6.3

Let us plug in the values in Henderson equation to find the ratio Base/Acid.

$7.5 = 6.3 + log \frac{[base]}{[acid]}$

$1.2 = log \frac{[base]}{[acid]}$

$\frac{[Base]}{[Acid]} = 10^{1.2}$

$\frac{[Base]}{[Acid]} = 15.8$

$[Base] = 15.8 \times [Acid]$

The total of mole fraction of acid and base is 1. Therefore we have,

$[Acid] + [Base] = 1$

But Base = 15.8 x [Acid]. Let us plug in this value in above equation.

$[Acid] + 15.8 \times [Acid] = 1$

$16.8 [Acid] = 1$

$[Acid] = \frac{1}{16.8}$

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The second option only.

LiOH, Li₂SO₄.

<h3>Explanation</h3>

A base neutralizes an acid when the two reacts to produce water and a salt.

Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:

Metal hydroxides

Ca(OH)₂ and
LiOH.

Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.

$\text{Ca}(\text{OH})_{\bf 2}+\text{H}_2\text{SO}_4 \to \textbf{Ca}\textbf{SO}_{\bf 4} +{\bf 2}\;\text{H}_2\text{O}$ .

The formula for calcium sulfate $\text{CaSO}_4$ in option A is spelled incorrectly. Why? The charge on each calcium $\text{Ca}^{2+}$ is +2. The charge on each sulfate ion ${\text{SO}_4}^{2-}$ is -2. Unlike $\text{Li}^{+}$ ions, it takes only one $\text{Ca}^{2+}$ ion to balance the charge on each ${\text{SO}_4}^{2-}$ ion. As a result, $\text{Ca}^{2+}$ and ${\text{SO}_4}^{2-}$ ions in calcium sulfate exist on a 1:1 ratio.

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Ammonia, NH₃

Ammonia NH₃ can also act as a base and neutralize acids. NH₃ exists as NH₄OH in water:

$\text{NH}_3 + \text{H}_2\text{O} \to \textbf{NH}_{\bf 4}\text{OH}$ .

The ion ${\text{NH}_4}^{+}$ acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt: